Evaluating a Complex Expression with Given Value

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SUMMARY

The expression $\large(1+x^5-x^7)^{{2014}^{1007}}$ can be evaluated by substituting $x=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}}{\sqrt{20}}$. This substitution simplifies the expression significantly, allowing for a straightforward calculation of the value. The discussion highlights contributions from members including Rido12 and Pranav, who provided correct solutions and methodologies for evaluating the expression.

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Evaluate $\large(1+x^5-x^7)^{{2014}^{1007}}$ if $x=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}}{\sqrt{20}}$.

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Congratulations to the following members for their correct solutions::)

1. Rido12
2. laura123
3. Olok
4. lfdahl
5. Pranav
6. kaliprasad

Solution from Rido12:
First, evaluate $x$:

$\begin{align*}x&=\dfrac{\sqrt{6+2\sqrt{5}}}{\sqrt{20}}+\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{20}}\\&=\sqrt{\dfrac{30+10\sqrt{5}}{100}}+\sqrt{\dfrac{30-10\sqrt{5}}{100}}\\&=\sqrt{\dfrac{25+5+10\sqrt{5}}{100}}+\sqrt{\dfrac{25+5-10\sqrt{5}}{100}}\\&=\frac{5+\sqrt{5}}{10}+\frac{5-\sqrt{5}}{10}\\&=1\end{align*}$Therefore,

$$\large(1+x^5-x^7)^{{2014}^{1007}}=\large(1)^{{2014}^{1007}}=1$$

Solution from Pranav;
$$x=\frac{\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}}{\sqrt{20}} \Rightarrow x^2=\frac{6+2\sqrt{5}+6-2\sqrt{5}+2\sqrt{36-10}}{20}=1$$
Hence,

$$\left(1-x^5+x^7\right)^{2014^{1007}}=\left(1-x^5\left(1-x^2\right)\right)^{2014^{1007}}=\boxed{1}$$
 

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